Proof.
Since X and Y are reduced and equidimensional of dimension 1, we see that Y is the scheme theoretic union of a subset of the irreducible components of X (in a reduced ring (0) is the intersection of the minimal primes). Let y \in Y be a closed point. If y is in the smooth locus of X \to \mathop{\mathrm{Spec}}(k), then y is on a unique irreducible component of X and we see that Y and X agree in an open neighbourhood of y. Hence Y \to \mathop{\mathrm{Spec}}(k) is smooth at y. If y is a node of X but still lies on a unique irreducible component of X, then y is a node on Y by the same argument. Suppose that y lies on more than 1 irreducible component of X. Since the number of geometric branches of X at y is 2 by Lemma 53.19.7, there can be at most 2 irreducible components passing through y by Properties, Lemma 28.15.5. If Y contains both of these, then again Y = X in an open neighbourhood of y and y is a node of Y. Finally, assume Y contains only one of the irreducible components. After replacing X by an open neighbourhood of x we may assume Y is one of the two irreducble components and Z is the other. By Properties, Lemma 28.15.5 again we see that X has two branches at y, i.e., the local ring \mathcal{O}_{X, y} has two branches and that these branches come from \mathcal{O}_{Y, y} and \mathcal{O}_{Z, y}. Write \mathcal{O}_{X, y}^\wedge \cong \kappa (y)[[u, v]]/(uv) as in Remark 53.19.9. The field \kappa (y) is finite separable over k by Lemma 53.19.7 for example. Thus, after possibly switching the roles of u and v, the completion of the map \mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y} corresponds to \kappa (y)[[u, v]]/(uv) \to \kappa (y)[[u]] and the completion of the map \mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y} corresponds to \kappa (y)[[u, v]]/(uv) \to \kappa (y)[[v]]. The scheme theoretic intersection of Y \cap Z is cut out by the sum of their ideas which in the completion is (u, v), i.e., the maximal ideal. Thus (2)(a) and (2)(b) are clear. Finally, (2)(c) holds: the completion of \mathcal{O}_{Y, y} is regular, hence \mathcal{O}_{Y, y} is regular (More on Algebra, Lemma 15.43.4) and \kappa (y)/k is separable, hence smoothness in an open neighbourhood by Algebra, Lemma 10.140.5.
\square
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